Given this reaction: $CH_4 + 2O_2 -> CO_2 + H_2O$, 16 g of $CH_4$ react with 64 g of $O_2$, producing 44 g of $CO_2$. How many grams of water are produced?

Md Ariful Islam

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Md Ariful Islam

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There are actually two methods of doing it - stoichimetry & also with the help of law of conservation of mass.

First by :

$16$grams of $CH_4$ means $1$mole of it.

Now, after balancing the entire equation we get:

$CH_4+2O_2 rarr CO_2+2H_2O$.

So, 1 mole of methane produces 2 moles of water. Hence, $36$ grams of water is formed.

Now by law of conservation of mass:

The entire mass of the reactants will be equal to the entire mass of the products after the reaction.

So, mass of water formed $=(16+64)-44$ grams, i.e. $36$ grams.

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